HDU 4417 Super Mario(離線線段樹or樹狀數組）
Problem Description
Mario is world-famous plumber. His “burly” figure and amazing jumping ability reminded in our memory. Now the poor princess is in trouble again and Mario needs to save his lover. We regard the road to the boss’s castle as a line (the length is n), on every
integer point i there is a brick on height hi. Now the question is how many bricks in [L, R] Mario can hit if the maximal height he can jump is H.
Input
The first line follows an integer T, the number of test data.
For each test data:
The first line contains two integers n, m (1 <= n <=10^5, 1 <= m <= 10^5), n is the length of the road, m is the number of queries.
Next line contains n integers, the height of each brick, the range is [0, 1000000000].
Next m lines, each line contains three integers L, R,H.( 0 <= L <= R < n 0 <= H <= 1000000000.)
Output
For each case, output "Case X: " (X is the case number starting from 1) followed by m lines, each line contains an integer. The ith integer is the number of bricks Mario can hit for the ith query.
Sample Input
1
10 10
0 5 2 7 5 4 3 8 7 7
2 8 6
3 5 0
1 3 1
1 9 4
0 1 0
3 5 5
5 5 1
4 6 3
1 5 7
5 7 3
Sample Output
Case 1:
4
0
0
3
1
2
0
1
5
1
普通的線段樹果斷T成狗，故轉向離線。
將磚塊高度val和瑪麗能跳的高度H都存起來，更新p[k].val<=q[i].h的，區間求和的時候就能保證當前更新的都是小於瑪麗能跳的的高度。。。
#include
#include
#include
#include
#include
typedef long long LL;
using namespace std;
const int maxn=1e5+10;
struct tree{
int l,r;
int cnt;
}t[maxn<<2];
struct node1{
int val;
int pos;
bool operator<(const node1 l1)const{
return val>1;
build(rs<<1,l,mid);
build(rs<<1|1,mid+1,r);
}
void update(int rs,int pos)
{
t[rs].cnt++;
if(t[rs].l==t[rs].r)
return ;
int mid=(t[rs].l+t[rs].r)>>1;
if(pos<=mid) update(rs<<1,pos);
else update(rs<<1|1,pos);
}
int query(int l,int r,int rs)
{
if(t[rs].l==l&&t[rs].r==r)
return t[rs].cnt;
int mid=(t[rs].l+t[rs].r)>>1;
if(r<=mid) return query(l,r,rs<<1);
else if(l>mid) return query(l,r,rs<<1|1);
else return query(l,mid,rs<<1)+query(mid+1,r,rs<<1|1);
}
int main()
{
int t,n,m;
int cas=1;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
build(1,1,n);
for(int i=1;i<=n;i++)
{
scanf("%d",&p[i].val);
p[i].pos=i;
}
for(int i=1;i<=m;i++)
{
scanf("%d%d%d",&q[i].l,&q[i].r,&q[i].h);
q[i].id=i;
}
sort(p+1,p+1+n);
sort(q+1,q+1+m);
int k=1;
for(int i=1;i<=m;i++)
{
while(k<=n&&p[k].val<=q[i].h)
{
update(1,p[k].pos);
k++;
}
ans[q[i].id]=query(q[i].l+1,q[i].r+1,1);
}
printf("Case %d:\n",cas++);
for(int i=1;i<=m;i++)
printf("%d\n",ans[i]);
}
return 0;
}
另外樹狀數組也可以做。類似的單點更新和區間求和問題都可以用樹狀數組來寫。
類似離線線段樹的思想將節點和詢問保存排序後依次更新，就能保證結果的正確性。
#include
#include
#include
#include
#include
typedef long long LL;
using namespace std;
const int maxn=1e5+10;
int cnt[maxn];
struct node1{
int val;
int pos;
bool operator<(const node1 l1)const{
return val0)
{
s+=cnt[x];
x-=lowbit(x);
}
return s;
}
int main()
{
int t,n,m;
int cas=1;
scanf("%d",&t);
while(t--)
{
memset(cnt,0,sizeof(cnt));
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
scanf("%d",&p[i].val);
p[i].pos=i;
}
for(int i=1;i<=m;i++)
{
scanf("%d%d%d",&q[i].l,&q[i].r,&q[i].h);
q[i].id=i;
}
sort(p+1,p+1+n);
sort(q+1,q+1+m);
int k=1;
for(int i=1;i<=m;i++)
{
while(k<=n&&p[k].val<=q[i].h)
{
update(p[k].pos);
k++;
}
ans[q[i].id]=query(q[i].r+1)-query(q[i].l);
}
printf("Case %d:\n",cas++);
for(int i=1;i<=m;i++)
printf("%d\n",ans[i]);
}
return 0;
}
